1 PRIORITY QUEUES AND HEAPS Lecture 19 CS2110 Spring 2014
Readings and Homework 2 Read Chapter 2 to learn about heaps Salespeople often make matrices that show all the great features of their product that the competitor s product lacks. Try this for a heap versus a BST. First, try and sell someone on a BST: List some desirable properties of a BST that a heap lacks. Now be the heap salesperson: List some good things about heaps that a BST lacks. Can you think of situations where you would favor one over the other? With ZipUltra heaps, you ve got it made in the shade my friend!
The Bag Interface 3 A Bag: interface Bag<E> { void insert(e obj); E extract(); //extract some element } boolean isempty(); Like a Set except that a value can be in it more than once. Example: a bag of coins Refinements of Bag: Stack, Queue, PriorityQueue
Stacks and Queues as Lists 4 Stack (LIFO) implemented as list insert(), extract() from front of list Queue (FIFO) implemented as list insert() on back of list, extract() from front of list All Bag operations are O(1) first 55 120 19 1 last
Priority Queue 5 A Bag in which data items are Comparable lesser elements (as determined by compareto()) have higher priority extract() returns the element with the highest priority = least in the compareto() ordering break ties arbitrarily
Priority Queue Examples Scheduling jobs to run on a computer default priority = arrival time priority can be changed by operator Scheduling events to be processed by an event handler priority = time of occurrence Airline check-in first class, business class, coach FIFO within each class
java.util.priorityqueue<e> 7 boolean add(e e) {...} //insert an element (insert) void clear() {...} //remove all elements E peek() {...} //return min element without removing //(null if empty) E poll() {...} //remove min element (extract) //(null if empty) int size() {...}
8 Priority Queues as Lists Maintain as unordered list insert() put new element at front O(1) extract() must search the list O(n) Maintain as ordered list insert() must search the list O(n) extract() get element at front O(1) In either case, O(n 2 ) to process n elements Can we do better?
9 Important Special Case Fixed number of priority levels 0,...,p 1 FIFO within each level Example: airline check-in insert() insert in appropriate queue O(1) extract() must find a nonempty queue O(p)
Heaps 10 A heap is a concrete data structure that can be used to implement priority queues Gives better complexity than either ordered or unordered list implementation: insert(): O(log n) extract(): O(log n) O(n log n) to process n elements Do not confuse with heap memory, where the Java virtual machine allocates space for objects different usage of the word heap
11 Heaps Binary tree with data at each node Satisfies the Heap Order Invariant: The least (highest priority) element of any subtree is found at the root of that subtree Size of the heap is fixed at n. (But can usually double n if heap fills up)
12 Heaps Smallest element in any subtree is always found at the root 4 of that subtree 14 21 8 19 35 22 38 55 10 20 Note: 19, 20 < 35: Smaller elements can be deeper in the tree!
13 Examples of Heaps Ages of people in family tree parent is always older than children, but you can have an uncle who is younger than you Salaries of employees of a company bosses generally make more than subordinates, but a VP in one subdivision may make less than a Project Supervisor in a different subdivision
Balanced Heaps 14 These add two restrictions: 1. Any node of depth < d 1 has exactly 2 children, where d is the height of the tree implies that any two maximal paths (path from a root to a leaf) are of length d or d 1, and the tree has at least 2 d nodes All maximal paths of length d are to the left of those of length d 1
Example of a Balanced Heap 15 4 14 21 8 19 35 22 38 55 10 20 d = 3
Store in an ArrayList or Vector 1 Elements of the heap are stored in the array in order, going across each level from left to right, top to bottom The children of the node at array index n are at indices 2n + 1 and 2n + 2 The parent of node n is node (n 1)/2
Store in an ArrayList or Vector 17 4 0 1 2 14 3 4 5 21 8 19 35 7 8 9 10 11 22 38 55 10 20 children of node n are found at 2n + 1 and 2n + 2
Store in an ArrayList or Vector 18 4 0 1 2 14 3 4 5 21 8 19 35 7 8 9 10 11 22 38 55 10 20 0 1 2 3 4 5 7 8 9 10 11 4 14 21 8 19 35 22 38 55 10 20 children of node n are found at 2n + 1 and 2n + 2
insert() 19 Put the new element at the end of the array If this violates heap order because it is smaller than its parent, swap it with its parent Continue swapping it up until it finds its rightful place The heap invariant is maintained!
insert() 20 4 14 21 8 19 35 22 38 55 10 20
insert() 21 4 14 21 8 19 35 22 38 55 10 20 5
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insert() 2 4 5 21 8 14 2 22 38 55 10 20 19 35
insert() 27 4 2 21 8 14 5 22 38 55 10 20 19 35
insert() 28 2 4 21 8 14 5 22 38 55 10 20 19 35
insert() 29 2 4 21 8 14 5 22 38 55 10 20 19 35
insert() 30 Time is O(log n), since the tree is balanced size of tree is exponential as a function of depth depth of tree is logarithmic as a function of size
insert() 31 /** An instance of a priority queue */ class PriorityQueue<E> extends java.util.vector<e> { /** Insert e into the priority queue */ public void insert(e e) { super.add(e); //add to end of array bubbleup(size() - 1); // given on next slide }
insert() 32 class PriorityQueue<E> extends java.util.vector<e> { /** Bubble element k up the tree */ private void bubbleup (int k) { int p= (k-1)/2; // p is the parent of k // inv: Every element satisfies the heap property except // element k might be smaller than its parent while (k > 0 && get(k).compareto(get(p)) < 0) { swap elements k and p; k= p; p= (k-1)/2; } }
extract() 33 Remove the least element it is at the root This leaves a hole at the root fill it in with the last element of the array If this violates heap order because the root element is too big, swap it down with the smaller of its children Continue swapping it down until it finds its rightful place The heap invariant is maintained!
extract() 34 4 5 21 8 14 35 22 38 55 10 20 19
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extract() 40 4 5 14 21 8 19 35 22 38 55 10 20
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extract() 42 4 5 14 21 8 19 35 22 38 55 10 20
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extract() 47 4 5 8 14 21 10 19 35 22 38 55 20
extract() 48 Time is O(log n), since the tree is balanced
49 /** Remove and return the smallest element return null if list is empty) */ public E extract() { } extract() if (size() == 0) return null; E temp= get(0); // smallest value is at root set(0, get(size() 1)); // move last element to the root setsize(size() - 1); // reduce size by 1 bubbledown(0); return temp;
50 /** Bubble the root down to its heap position. Pre: tree is a heap except: root may be >than a child */ private void bubbledown() { int k= 0; // Set c to smaller of k s children int c= 2*k + 2; // k s right child if (c > size()-1 get(c-1).compareto(get(c)) < 0) c= c-1; // inv tree is a heap except: element k may be > than a child. // Also. k s smallest child is element c while (c < size() && get(k).compareto(get(c) > 0) { Swap elements at k and c; k= c; c= 2*k + 2; // k s right child if (c > size()-1 get(c-1).compareto(get(c)) < 0) c= c-1; } }
51 HeapSort Given a Comparable[] array of length n, Put all n elements into a heap O(n log n) Repeatedly get the min O(n log n) public static void heapsort(comparable[] b) { PriorityQueue<Comparable> pq= new PriorityQueue<Comparable>(b); for (int i = 0; i < b.length; i++) { b[i] = pq.extract(); } } One can do the two stages in the array itself, in place, so algorithm takes O(1) space.
PQ Application: Simulation 52 Example: Probabilistic model of bank-customer arrival times and transaction times, how many tellers are needed? Assume we have a way to generate random inter-arrival times Assume we have a way to generate transaction times Can simulate the bank to get some idea of how long customers must wait Time-Driven Simulation Check at each tick to see if any event occurs Event-Driven Simulation Advance clock to next event, skipping intervening ticks This uses a PQ!